nindita
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how to obtain delta H
for example a liquid toluen will be heated from 300 K to 500 K
i ve data Cp liquid for toluene at 300 K and Cp liquid at 500K
question is how to calculate the deltaH
deltaH= integral Cp dt??
howw???
thx in advance..
PS. I dont have yaws... 
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guy
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deltaH of what? Formation, vaporization?
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Nerro
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I don't mean to whine but aren't Cp and Cv values only for gases? What is "Cp liquid"?
Anyway, (delta = /\),
/\H = /\e + p /\V (H = enthalpy, p = pressure, V = volume)
/\e = Cv/\T
Cp/\T = Cv/\T + R /\T and knowing that pV = RT => Cp/\T = Cv/\T + nRT
Also Cp = Cv + nR and gamma = Cp / Cv the values for gamma may be found in special tables.
Does that help you any? I have a hunch that this is a homework assignment and I don't intend to tell you everything.
#261501 +(11351)- [X]
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Engager
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| Quote: | Originally posted by nindita
for example a liquid toluen will be heated from 300 K to 500 K
i ve data Cp liquid for toluene at 300 K and Cp liquid at 500K
question is how to calculate the deltaH
deltaH= integral Cp dt??
howw???
thx in advance..
PS. I dont have yaws... |
Simple, just calculate the interpolation function from points given in books, and after that integrate it in bounds needed. In books for many
substances there are equitions to their isobaric heat campacity in form like Cp = a + bT + c/(T^2), where a, b and c are some numbers. Those equitions
are made by fitting of experimentaly determined values on some temperature points. Int is integral, (a,b) - bounds of integral, [] - eq. to
integration.
Heat of formation is: H = Int [Cp*dT]
Entropy is: S = Int [(Cp/T)*dT]
Target temperature where we need Hf calculated is T.
So we have:
Hf(T) = Hf(298) + Int(T,298)[Cp*dT]
S(T) = S(298) + Int(T,298)[(Cp/T)*dT]
Placing Cp function to this equitions gives:
Int(T,298)[Cp*dT] = Int(T,298) [a+bT+c/(T^2)] = (T,298)(aT + b(T^2)/2 - c/T) = a(T-298) + b/2((T^2)-298^2) - c (1/T - 1/298))
Int(T,298)[(Cp/T)*dT] = Int(T,298) [a/T + b + c/(T^3)] = (T,298)(alnT + bT - 2c/(T^2)) = a ln (T/298) + b(T-298) - 2c (1/(T^2)+1/(298^2))
Finaly we get:
Hf(T) = H(298) + a(T-298) + b/2((T^2)-298^2) - c (1/T - 1/298))
S(T) = S(298) + a ln (T/298) + b(T-298) - 2c (1/(T^2)+1/(298^2))
Also you must remember that if there are some phase transitions on the intervall of (T,298) you must add them to Hf(T) and S(T). Remember that such Cp
equitions are valid only in defined temperature intervals (usualy bounded by phase thansition tempereatures or standart 298K).
So for example for water we need to calculate Hf at 300C = 573K. From book we take to equitions Cp1=f(T) [273,373], Cp2=f(T) [373, x (x>573K)],
Hf(298), S(298), heat of evaporation (Hevap) and entropy of evaporation (Sevap). 373K is a water boiling point (phase transition point).
So:
Hf(573K) = H(298) + Int(373,298)(Cp1*dT) + Hevap + Int(573,373)(Cp2*dT)
S(573K) = S(298) + Int(373,298)((Cp1/T)*dT) + Sevap + Int(573,373)((Cp2/T)*dT)
In such manner you can integrate any Cp equition, or integrate Cp equition made by interpolation of experimental points. I'll think that helps.
[Edited on 24-8-2006 by Engager]
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Quibbler
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If you don't want to make it that difficult you can just take the average of Cp at 300&500K (often the 'c' term is small).
so /\H (500K) = /\H (300K)+Cp(average)*(500-300).
Oh and as liquid are almost incompressable then Cp=Cv as the PdV term is zero.
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